Question

por favor resolvam a 3, a e b!!

Answer 1

a) $A=\dfrac{\mathrm{sen\,}0+\mathrm{sen\,}^{\pi}\!\!\diagup\!\!_{2}-\mathrm{sen\,}^{\pi}\!\!\diagup\!\!_{6}}{\mathrm{sen\,}^{7\pi}\!\!\diagup\!\!_{6}}$

$A=\dfrac{0+1-^{1}\!\!\diagup\!\!_{2}}{-\mathrm{sen}\left(^{7\pi}\!\!\diagup\!\!_{6}-\pi \right )}\\ \\ A=\dfrac{^{2}\!\!\diagup\!\!_{2}-^{1}\!\!\diagup\!\!_{2}}{-\mathrm{sen}\left(^{7\pi}\!\!\diagup\!\!_{6}-^{6\pi}\!\!\diagup\!\!_{6} \right )}\\ \\ A=\dfrac{^{1}\!\!\diagup\!\!_{2}}{-\mathrm{sen\,}^{\pi}\!\!\diagup\!\!_{6}}\\ \\ A=\dfrac{^{1}\!\!\diagup\!\!_{2}}{-^{1}\!\!\diagup\!\!_{2}}\\ \\ \boxed{A=-1}$


b) $B=\dfrac{\mathrm{sen\,}^{3\pi}\!\!\diagup\!\!_{2}-\mathrm{sen\,}\pi+\mathrm{sen\,}^{\pi}\!\!\diagup\!\!_{6}}{2\cdot \mathrm{sen\,}^{5\pi}\!\!\diagup\!\!_{6}+\mathrm{sen\,}0}$

$B=\dfrac{-1-0+^{1}\!\!\diagup\!\!_{2}}{2\cdot \mathrm{sen}\left(\pi-^{5\pi}\!\!\diagup\!\!_{6} \right )+0}\\ \\ B=\dfrac{^{-2}\!\!\diagup\!\!_{2}+^{1}\!\!\diagup\!\!_{2}}{2\cdot \mathrm{sen}\left(^{6\pi}\!\!\diagup\!\!_{6}-^{5\pi}\!\!\diagup\!\!_{6} \right )}\\ \\ B=\dfrac{^{-1}\!\!\diagup\!\!_{2}}{2\cdot \mathrm{sen}\left(^{\pi}\!\!\diagup\!\!_{6} \right )}\\ \\ B=\dfrac{^{-1}\!\!\diagup\!\!_{2}}{2\cdot ^{1}\!\!\diagup\!\!_{2} \right )}\\ \\ B=\dfrac{^{-1}\!\!\diagup\!\!_{2}}{1}\\ \\ \boxed{B=-^{1}\!\!\diagup\!\!_{2}}$